(a) The object moves with uniform velocity from A to B.
(b) The object moves with constant velocity from B to C.
(c) The object moves with increasing velocity from C to D.
<h3>
Velocity of the object from point A to B</h3>
V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s
<h3>
Velocity of the object from point B to C</h3>
V(B to C) = (6 - 6)/(11 - 4) = 0 m/s
<h3>
Velocity of the object from point C to D</h3>
V(C to D) = (7 - 6)/(12 - 11) = 1 m/s
final velocity = 1 + 1.5 m/s = 2.5 m/s
Thus, we can conclude the following;
The object moves with uniform velocity from A to B.
The object moves with constant velocity from B to C.
The object moves with increasing velocity from C to D.
Learn more about velocity here: brainly.com/question/6504879
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The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
I'd say move faster, unless it's asking something else.
Answer:
In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .
Answer:
Option B
Explanation:
Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.
For galvanometer B, the reading is 4 mA.
While for galvanometer C, the reading is -5 MA
Thus, option B is correct.
